Question

Two identical parallel plate capacitors, of capacitance $C$ each, have plates of area $A$, separated by a distance $d$. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants $K_1, K_2$ and $K_3$. The first capacitor is filled as shown in fig. $I$, and the second one is filled as shown in fig. $II$. If these two modified capacitors are charged by the same potential $V$, the ratio of the energy stored in the two, would be ($E_1$ refers to capacitor (I) and $E_2$ to capacitor (II) :

• A. $\frac{E_{1}}{E_{2}} = \frac{9K_{1}K_{2}K_{3}}{\left(K_{1} +K_{2}+K_{3}\right)\left(K_{2}K_{3} + K_{3}K_{1} +K_{1}K_{2}\right)} $
• B. $\frac{E_{1}}{E_{2}} = \frac{K_{1}K_{2}K_{3}}{\left(K_{1} +K_{2}+K_{3}\right)\left(K_{2}K_{3} + K_{3}K_{1} +K_{1}K_{2}\right)} $
• C. $\frac{E_{1}}{E_{2}} = \frac{\left(K_{1} +K_{2}+K_{3}\right)\left(K_{2}K_{3} + K_{3}K_{1} +K_{1}K_{2}\right)}{K_{1}K_{2}K_{3}} $
• D. $\frac{E_{1}}{E_{2}} = \frac{\left(K_{1} +K_{2}+K_{3}\right)\left(K_{2}K_{3} + K_{3}K_{1} +K_{1}K_{2}\right)}{9K_{1}K_{2}K_{3}} $

Answer 1

The Correct Option is D

To solve for the ratio of energy stored in two different configurations of parallel plate capacitors, we need to understand how dielectric materials affect capacitance. The dielectric materials change the capacitance based on their dielectric constant when inserted between capacitor plates.

**Step 1: Determine Capacitance for Both Configurations**

**Capacitor (I):**

The dielectric materials in the first capacitor configuration divide the gap between the plates such that each dielectric occupies equal thickness d/3 of the total separation d between the plates. The equivalent capacitance C_1 is given by the series combination formula:

C_1 = \left(\frac{1}{C_{1a}} + \frac{1}{C_{1b}} + \frac{1}{C_{1c}}\right)^{-1}

where C_{1a} = \frac{K_1 \cdot \epsilon_0 \cdot A}{d/3}, C_{1b} = \frac{K_2 \cdot \epsilon_0 \cdot A}{d/3}, and C_{1c} = \frac{K_3 \cdot \epsilon_0 \cdot A}{d/3}.

This simplifies to:

\frac{1}{C_1} = \frac{3}{K_1 \cdot \epsilon_0 \cdot A} + \frac{3}{K_2 \cdot \epsilon_0 \cdot A} + \frac{3}{K_3 \cdot \epsilon_0 \cdot A}

Hence, take the reciprocal to find C_1:

C_1 = \frac{\epsilon_0 \cdot A \cdot (K_1 \cdot K_2 \cdot K_3)}{3 \cdot (K_2 \cdot K_3 + K_3 \cdot K_1 + K_1 \cdot K_2)}

**Capacitor (II):**

For the second capacitor, the dielectrics are in parallel and occupy the entire plate area. Thus, the equivalent capacitance C_2 is the sum of each capacitance:

C_2 = \frac{K_1 \cdot \epsilon_0 \cdot A}{d} + \frac{K_2 \cdot \epsilon_0 \cdot A}{d} + \frac{K_3 \cdot \epsilon_0 \cdot A}{d}

Thus simplifying:

C_2 = \frac{\epsilon_0 \cdot A \cdot (K_1 + K_2 + K_3)}{d}

**Step 2: Energy Stored in Each Capacitor**

The energy stored in a capacitor is given by:

E = \frac{1}{2} C V^2

Thus:

E_1 = \frac{1}{2} \cdot C_1 \cdot V^2 \quad \text{and} \quad E_2 = \frac{1}{2} \cdot C_2 \cdot V^2

The ratio is:

\frac{E_1}{E_2} = \frac{C_1}{C_2} = \frac{(K_1 \cdot K_2 \cdot K_3)}{3 (K_2 \cdot K_3 + K_3 \cdot K_1 + K_1 \cdot K_2)} \times \frac{d}{(K_1 + K_2 + K_3)}

Simplifying further:

\frac{E_1}{E_2} = \frac{(K_1 + K_2 + K_3)(K_2K_3 + K_3K_1 + K_1K_2)}{9K_1K_2K_3}

This matches the provided correct answer.